What is the derivative of 2tsin(t) ?

The derivative of 2tsin(t) is 2(sin(t)+tcos(t))

What is the first derivative of 2tsin(t) ?

The first derivative of 2tsin(t) is 2(sin(t)+tcos(t))

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derivative of

2 t sin (t)

2 (sin (t) + t cos (t))

Solution steps

\frac{d}{d t} (2 t sin (t))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d t} (t sin (t))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = sin (t)

= 2 (\frac{d t}{d t} sin (t) + \frac{d}{d t} (sin (t)) t)

\frac{d t}{d t} = 1

\frac{d}{d t} (sin (t)) = cos (t)

= 2 (1 \cdot sin (t) + cos (t) t)

Simplify

= 2 (sin (t) + t cos (t))

\frac{d}{d x} (\frac{3 x ^{2} + 2 y ^{2}}{( x - 2 y ) ^{2}})

f (x) = x, at x = \frac{1}{9}

\frac{d}{d x} (sin (2 x))

\frac{\partial}{\partial x} (5 x^{\frac{4}{5}} y^{\frac{1}{5}})

\frac{d}{d x} (cos ((π x)^{7}))

What is the derivative of 2tsin(t) ?
The derivative of 2tsin(t) is 2(sin(t)+tcos(t))
What is the first derivative of 2tsin(t) ?
The first derivative of 2tsin(t) is 2(sin(t)+tcos(t))